We have the following ODE: $$x'(t) = f(t,x) \quad \wedge \quad x(t_0) = x_0$$
Let's say we have a linear multistep ($q$ steps) method:
$$\sum_{j=0}^q \alpha_j x_{k+j} = h \sum_{j=0}^q \beta_j f_{k+j}$$
It's local truncation error is equal to:
$$\sum_{j=0}^q (\alpha_j x(t_k + h j) -h \beta_j x' (t_k + hj))$$
If we assume that $x \in \pi_3$, this means, that:
$$x(t) = {\gamma}_0 + {\gamma}_1 t + ... + {\gamma}_p t^p$$
$$x'(t) = {\gamma}_1 + 2{\gamma}_2 t + ... + p {\gamma}_p t^{p-1}$$
This means, that:
$$\alpha_j x(t_k + h j) = \alpha_j{\gamma}_0 + \alpha_j{\gamma}_1 (t_k + h j) + ... + \alpha_j{\gamma}_p (t_k + h j)^p$$
$$-h \beta_j x'(t_k + h j) = -h \beta_j{\gamma}_1 - 2h \beta_j{\gamma}_2 (t_k + h j) - ... - p h \beta_j{\gamma}_p (t_k + h j)^{p-1}$$
Now let's add these two together, and we get:
$$\sum_{j=0}^q ((\alpha_j {\gamma}_0 - h \beta_j {\gamma}_1) + (\alpha_j {\gamma}_1 - 2h \beta_j {\gamma}_2)(t_k + hj) + ... + (\alpha_j {\gamma}_{p-1} - p h \beta_j {\gamma}_p)(t_k + hj)^{p-1} + (\alpha_j {\gamma}_p) (t_k + hj)^p)$$
This must be equal to zero. This means, that:
$$\sum_{j=0}^q ({\gamma}_p \alpha_j) = 0$$
$$\sum_{j=0}^q ({\gamma}_{p-1} \alpha_j - p h \beta_j {\gamma}_p) = 0$$
$$...$$
$$\sum_{j=0}^q ({\gamma}_0 \alpha_j - h \beta_j {\gamma}_1) = 0$$
Because $\sum_{j=0}^q ({\gamma}_{p-1} \alpha_j - p h \beta_j {\gamma}_p) = 0$, ..., $\sum_{j=0}^q ({\gamma}_0 \alpha_j - h \beta_j {\gamma}_1) = 0$, we can write this shorter as:
$$\sum_{j=0}^q ({\gamma}_p \alpha_j) = 0$$
$$\sum_{k=1}^p \sum_{j=0}^q ({\gamma}_{k-1} \alpha_j - h \beta_j {\gamma}_k) = 0$$
Now it get's to the point where my attempt to prove this theorem fails:
There is a theorem, that if we have a linear multistep (of $q$ steps) method, then it is of rank $p$, if:
$$c_0 = \sum_{j=0}^q \alpha_j = 0$$
And for all $k = 1, ..., p$:
$$c_k = \sum_{j=0}^q (\frac{1}{k!} j^k \alpha_j - \frac{1}{(k-1)!} j^{k-1} \beta_j) = 0$$
And $c_{p+1} \neq 0$
Again, instead of $\forall$ we can write it shorter:
$$\sum_{j=0}^q \alpha_j = 0$$
$$\sum_{k=1}^p \sum_{j=0}^q (\frac{1}{k!} j^k \alpha_j - \frac{1}{(k-1)!} j^{k-1} \beta_j) = 0$$
Now let's compare it to what we got previously:
$$\sum_{j=0}^q ({\gamma}_p \alpha_j) = 0$$
$$\sum_{k=1}^p \sum_{j=0}^q ({\gamma}_{k-1} \alpha_j - h \beta_j {\gamma}_k) = 0$$
It's pretty easy to see, that if $\sum_{j=0}^q \alpha_j = 0$, then $\sum_{j=0}^q ({\gamma}_p \alpha_j) = 0$ and vice versa (we either multiply oth sides by $\gamma_p$, or divide both sides).
But what about $\sum_{k=1}^p \sum_{j=0}^q ({\gamma}_{k-1} \alpha_j - h \beta_j {\gamma}_k) = 0$? That's where I am stuck
Any ideas what I could do?